${{n}_{{{C}_{2}}{{H}_{2}}}}=\frac{{{V}_{{{C}_{2}}{{H}_{2}}}}}{22,4}=\frac{5,6}{22,4}=0,25\left( mol \right)$

${{C}_{2}}{{H}_{2}}+2B{{r}_{2}}\to {{C}_{2}}{{H}_{2}}B{{r}_{4}}$

 0,25 $\to $  0,5                              (mol)

Theo phương trình ta có:

${{n}_{B{{r}_{2}}}}=2.{{n}_{{{C}_{2}}{{H}_{2}}}}=2.0,25=0,5\left( mol \right)$

$\Rightarrow {{m}_{B{{r}_{2}}}}={{n}_{B{{r}_{2}}}}.{{M}_{B{{r}_{2}}}}=0,5.160=80\left( g \right)$

$C%=\frac{{{m}_{B{{r}_{2}}}}}{{{m}_{\text{d}{{\text{d}}_{B{{r}_{2}}}}}}}.100=\frac{80}{500}.100=16\left( % \right)$

$\Rightarrow $Chọn đáp án D